Question

Determine the zeros of the given polynominal function and the multiplicity of each zero. List the zeros from smallest to largest. If the zero is not an integer then use a fraction not a decimal.

Answer 1

So we have to find the zeros of a polynomial and their multiplicity. The zeros of a function are the x values that make the function equal to 0. In this case the function is a polynomial:

$G(x)=3x^4+6x^3+3x^2$

It is important to note that polynomials can be factored like:

$f(x)=A(x-r_1)^a(x-r_2)^b\cdot\cdot\cdot(x-r_n)^m$

Where the r's are the different zeros of the polynomial, A is the leading coefficient (the number multiplying the greatest power of x) and the exponents of each term are the multiplicity of the correspondant zero.

So let's look for the zeros of G(x) by equalizing it to 0:

$3x^4+6x^3+3x^2=0$

It is important to note that each of the terms in the left side can be written as the term 3x^2 multiplied by another term:

$\begin{gathered} 3x^4+6x^3+3x^2=3x^2\cdot x^2+3x^2\cdot2x+3x^2\cdot1=0 \\ 3x^2\cdot(x^2+2x+1)=0 \end{gathered}$

We can write x as x-0 and we have:

$3(x-0)^2(x^2+2x+1)=0$

So the solutions to this equation are the values of x that make the expression on the left equal to zero. This expression is composed of a constant multiplied by two terms inside parenthesis. This means that the expression is equal to 0 when any of those two terms inside parenthesis is equal to 0. If x=0 then the left term is equal to 0 so we already have one solution: x=0. In order to find the x values that make the right term equal to 0 we have to solve:

$x^2+2x+1=0$

This is a quadratic expression equal to 0. This means that we can use the quadratic solving formula to find its solutions. Remember that for an equation:

$ax^2+bx+c=0$

The solving formula is:

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

In our case a=1, b=2 and c=1 so we have:

$x=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot1}}{2\cdot1}=\frac{-2\pm\sqrt[]{4-4}}{2}=\frac{-2\pm0}{2}$

Then the two values of x are:

$\begin{gathered} x=\frac{-2+0}{2}=-1 \\ x=\frac{-2-0}{2}=-1 \end{gathered}$

The values are the same: x=-1. This means that the quadratic term inside parenthesis is equal to 0 when x=-1. Then it can be rewritten as:

$x^2+2x+1=(x-(-1))(x-(-1))=(x+1)^2$

Now we can go back to the expression of G(x)=0:

$G(x)=3(x-0)^2(x+1)^2=0$

So we have found the factored form of G(x):

$3(x-0)^2(x+1)^2$

Which means that its zeros are -1 and 0 and both have a multiplicity of 2 because their terms are squared. Then the answers are:

1st zero = -1

with a multiplicity of 2

2nd zero= 0

with a multiplicity of 2